* Corresponding Author DOI: 10.37094/adyujsci.594199
Inverse Spectral Problems for Second Order Difference Equations with
Generalized Function Potentials by aid of Parseval Formula
Bayram BALA1,_{*}
1_{Harran University, Department of Mathematics, Şanlıurfa, Turkey. }
[email protected], ORCID: 0000000165156039
Received: 19.07.2019 Accepted: 03.03.2020 Published: 25.06.2020
Abstract
In the present study we are investigated inverse spectral problems for spectral analysis and two spectra of matrix by using equality which is equivalance Parseval formula. The matrix is tridiagonal almostsymmetric matrix. The mean of almostsymmetric is the entries above and below the main diagonal are the same except the entries and .
Keywords: Parseval formula; Spectral analysis; Two spectra.
Parseval Formülü yardımıyla Genelleşmiş Fonksiyon Katsayılı İkinci Mertebeden Fark Denklemleri için Ters Spektral Problemler
Öz
Bu çalışmada, Parseval formülü ile eşdeğer olan eşitlik kullanılarak matrisinin spektral analizine göre ve iki spektrumuna göre ters spektral problemleri incelenmiştir. , tipinde hemen hemen simetrik üçköşegensel matristir. Hemen hemen simetriklik, ve elemanları dışında matrisin köşegeninin altında ve üstündeki elemanları eşit olmasıdır.
J J N N´ M
a
c
_{M} J J N N´ Ma
c
_{M}Anahtar Kelimeler: Parseval formülü; Spektral analiz; İki spektrum. 1. Introduction
Consider the second order difference equation
(1) with the boundary conditions
(2)
where is column vector which is solution of the second order difference equation, is constant
𝜌!= #
1, 0 ≤ 𝑛 ≤ 𝑀
𝛼, 𝑀 < 𝑛 ≤ 𝑁 − 1 , 𝛼 ∈ ℝ
"_{− {1}, (3) }
and 𝑎_{!}, 𝑏_{!}∈ ℝ, 𝑎_{!}> 0 are coefficients of Eqn. (1),
(4)
Now, we can write the Eqn. (1) by definition of
is tridiagonal almostsymmetric matrix and the entries of are the coefficients of Eqn. (1).
{
}
1 1 1,
1 11,
0,1,...,
1 ,
n n n n n n n n Na y
_{} _{}+
b y
+
a y
_{+}=
lr
y
a
_{}=
c
_{}=
n
Î
N
1 N0,
y
_{}=
y
=
{ }
1 0 N n ny
=
y
_{=} nr
{
}
{
}
,
,
1,...,
2 ,
,
1,
2,...,
1 .
n n n nc
a
n
M M
N
d
b
n
M
M
N
a
a
=
Î
+
=
Î
+
+
nr
{
}
{
}
1 1 1 1 1 1,
0,1,...,
,
,
1,
2,...,
1 .
n n n n n n n n n n n n n na y
b y
a y
y
n
M
c y
d y
c y
y
n
M
M
N
l
l
  +   ++
+
=
Î
ìï
í
+
+
=
Î
+
+
ïî
JN N
´
J.
So, the eigenvalue problem is equivalent problem (1)(3) which is discrete form SturmLiouville problem with discontinuous coefficients
(5)
where is a piecewise function
H. Hochstadt made significant contributions to the development of the inverse problem for second order difference equations. He studied inverse problem for Jacobi matrices in [14]. Later, G. Guseinov has pioneered for inverse problem of infinite symmetric tridiagonal matrices. He considered different kinds of inverse spectral problems for second order difference equation; such as the inverse spectral problems of spectral analysis for infinite Jacobi matrices in [5], the inverse spectral problems for the infinite nonself adjoint Jacobi matrices from generalized spectral function in [6, 7], and the inverse spectral problems for same matrices from spectral data and two spectra in [812]. The inverse spectral problem for discrete form of SturmLiouville problem with continuous coefficients has been studied in [13] and the inverse spectral problem with spectral parameter in the initial conditions has been studied by M. Manafov in [14]. The eigenvalues and eigenfunctions and the inverse problem for SturmLiouville operator with discontinuous coefficients which is the same problem given by (5) are investigated by E. Akhmedova and H.
0 0 0 1 1 1 2 1 1 1 1 1 1 2 3 3 3 2 2 2 1
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
M M M M M M M M M M N N N N N N Nb
a
a
b
a
a
b
b
a
a
b
a
J
c
d
c
c
d
d
c
c
d
c
c
d
 + + + +    é
=
ë
!
!
!
!
!
!
"
"
" #
"
"
"
"
#
"
"
"
!
!
!
!
!
!
!
!
"
"
" #
"
"
"
"
#
"
"
"
!
!
!
!
!
!
ù
ê
ú
ê
ú
ê
ú
ê
ú
ê
ú
ê
ú
ê
ú
ê
ú
ê
ú
ê
ú
ê
ú
ê
ú
ê
ú
ê
ú
ê
ú
ê
ú
ê
_{úû}
Jy
=
l
y
( )
( )
( ) ( )
( ) ( )
[ ]
( )
( )
,
, ,
0,
d
d
p x
y x
q x y x
x y x
x
a b
dx
dx
y a
y b
lr
é
_{ù +}
_{=}
_{Î}
ê
ú
ë
û
=
=
( )
x
r
( )
2 21,
,
1.
, c
a x c
x
x b
r
a
a
£ £
ì
=
_{í}
¹
< £
î
Huseynov in [15, 16], respectively. Bala et al. are studied inverse spectral problem for almost symmetric tridiagonal matrices from generalized spectral function in [17] and they examined inverse spectral problems for same matrices from spectral data and two spectra in [18]. Finite dimensional inverse problems are investigated by H. Huseynov in [19].
Also, Parseval equality of discrete SturmLiouville equation with periodic generalized function potentials is studied by Manafov et al. in [20]. At the same time a new approach for higherorder difference equations and eigenvalue problems is examined by Bas and Ozarslan in [21].
The goal of this article is to study inverse spectral problems of the problem (1)(3) for spectral analysis and two spectra by using Parseval formula.
2. Direct Problem for Spectral Analysis
The matrix has number eigenvalues and number eigenvectors , which form an orthonormalized basis. Assume that the eigenvalues are real. We bring to mind the algorithm of structure for the matrix eigenvalues and eigenvectors.
Let be a solution of Eqn. (1)
(6) with initial conditions
, (7)
and the degree of polynomial is .
Lemma 1. The following equality holds:
.
Proof. See [17].
According to Lemma 1, the roots of the equation are equal the eigenvalues of , and eigenvectors corresponding eigenvalues will be
J N
l l
_{1}, ,...,
_{2}l
_{N} N 1, ,...,
2 Nv v
v
J( )
nP
l
( )
( )
( )
( )
{
}
1 1 1,
0,1,...,
1 ,
n n n n n n n na P
_{} _{}l
+
b P
l
+
a P
_{+}l lr
=
P
l
n
Î
N
( )
( )
10,
01
P
_{}l
=
P
l
=
( )
nP
l
n
(
) ( )
0 1 1 1( )
det
J

l
I
= 
1
Na a a c
...
_{M M}_{+}...
c P
_{N}_{} _{N}l
( )
nP
l
J,
1,
kk
N
l
=
.
Assuming that , where . Thus, we have the complete
orthonormalized system of eigenvectors of the matrix . The numbers are called normalized numbers of the problem (1)(3).
Lemma 2. Eigenvalues of matrix are different.
Proof. Because of eigenvalues are the roots of polynomial , we must show that . Firstly, take the derivative equation Eqn. (6) by , we have
(8)
Now, if the Eqn. (8) is multiplied by and the Eqn. (6) is multiplied by , the second result is substracted from the first, for we obtain
. (9)
For and summing from to , pay attention to Eqn. (7) and we have
. (10)
As a result, .
According to Lemma 2, we can assume that . The following Lemma is about Parseval equality.
Lemma 3. The expansion formula which is equivalent Parceval equality, can be written as
below: , (11)
( )
(
0( ) ( )
,
1,...,
1( )
)
T kP
kP
kP
N kl
l
l
_{}l
Â
=
( )
k k kv
l
b
Â
=
1 2( )
0 N k j k jP
b
l
==
å
Jb
_{k} J,
1, 2,...,
kk
N
l
=
P
_{N}( )
l
( )
0
N kP
¢
l
¹
l
( )
( )
( )
( )
( )
1 1 1.
n n n n n n n n n na P
_{}¢
_{}l
+
b P
¢
l
+
a P
¢
_{+}l lr
=
P
¢
l r
+
P
l
( )
nP
l
P
_{n}¢
( )
l
{
0,1,...,
1
}
n
Î
N
( ) ( )
( ) ( )
(
)
(
( ) ( )
( ) ( )
)
2( )
1 1 1 1 1 n n n n n n n n n n n na
_{}P
¢
_{}l
P
l

P
¢
l
P
_{}l

a P
¢
l
P
_{+}l

P
¢
_{+}l
P
l
=
r
P
l
kl l
=
n
0 N1P
_{N}( )
l
_{k}=
0
( )
( )
2( )
1 2( )
1 1 0 1 M N N N k N k j j j j Ma P
_{}l
P
_{}l
P
l
a
P
l
= = +¢
=
å
+
å
( )
0
N kP
¢
l
¹
1 2...
Nl l
<
< <
l
( ) ( )
1,
,
0,
1
N m j n j mn j jP
P
m n
N
h
_{l}
_{l}
_{d}
b
==
=
å
where is defined by
, (12)
and is the Kronecker delta.
For in the Eqn. (11) and from conditions (7) we obtain following equality
. (13)
Thus, we get eigenvalues and eigenvectors corresponding . So, we can say that the direct spectral problem of spectral analysis is solved.
Now let's try to answer the following question:
If we know eigenvalues and eigenvectors of matrix , is it possible to reconstruct the matrix by using the following formula
,
where scalar product.
It is clear that eigenvalues of is not sufficient for reconstruct matrix . On account of this we must have some more information about eigenvectors.
Definition 4. The collection of quantitites are called spectral data for the matrix
Additional we will need the presentation of entries of the matrix by the polynomial . For the Eqn. (6) is multiplied by , then summing by from to and using Lemma 3, we have
h
1,
or
, or
m
n M
m
n M
h
a
£
ì
= í
_{>}
î
mnd
0 n m= = 11
1
N j=b
j=
å
{ }
N_{1} k kl
_{=}v
_{j},
j
=
1, 2,...,
N
{ }
N_{1} k kl
_{=}{ }
N_{1} k kl
_{=}{ }
N_{1} k kv
_{=} J J(
)
2(
)
1 , , 0, 1 N k k k k Jul
u v v u l N = =å
Î( )
1 0,
N _{j j} ju v
u v
==
å
J J{
l b
_{k},
_{k}}
. J J( )
nP
l
l l
=
_{j} _{m}( )
_{j} jP
h
_{l}
b
j
1
N
where is defined by (12). It is clear that for . Then, we can write these equalities as below:
(14)
(15)
(16)
(17)
(18)
3. Inverse Problem of Spectral Analysis The inverse problem of spectral analysis is reconstruct matrix by using the collection quantities . Theorem 5. Let an arbitrary collection of matrix is . In order for this collection to be spectral data for some matrix which have form , it is necessary and sufficient that the following conditions are satisfied: (i) , (ii) , (iii) .
( ) ( )
2 1 1,
0, N 2 M,
N j n n j n j j ja
h l
P
l
P
l
n
b
+ ==
å
=
( )
2 2 1,
0, N 1,
N j n n j j jb
h l
P
l
n
b
==
å
=
h
r h a
_{n}= =
m
or
n M
>
( ) ( )
1 1,
0,
1,
N j n n j n j j ja
l
P
l
P
l
n
M
b
+ ==
å
=
( )
1( )
( )
1( )
1 1,
,
N N j j M M j M j M M j M j j j j ja
al
P
l
P
l
c
l
P
l
P
l
b
+b
+ = ==
å
=
å
( ) ( )
1 1,
1,
2,
N j n n j n j j jc
al
P
l
P
l
n M
N
b
+ ==
å
=
+
( )
2 1,
0,
,
N j n n j j jb
l
P
l
n
M
b
==
å
=
( )
2 1,
1,
1.
N j n n j j jd
al
P
l
n M
N
b
==
å
=
+
J{
l b
_{k},
_{k}}
{
l b
_{k},
_{k}}
J J k jl
¹
l
11
1
N j=b
j=
å
a
_{n}>
0,
n
=
0,
N

2
Lemma 6. Let are distinct real numbers and for the positive numbers be given that . Then there exists unique polynomials
with and positive leading coefficients satisfying the conditions (11).
Now, we will give another method for a kind of approach to the solution of the inverse spectral problem which is called the GelphandLevitanMarchenko method.
Let be a solution of the Eqn. (1) satisfying the conditions
,
in the case .
Recall that is a polynomial of degree , so it can be expressed as
, (19)
where and are constants. There is a connection between coefficients and .
Then we can write the equalities
(20)
(21)
Now, we can write from Eqn. (19)
,
1,
kk
N
l
=
,
1,
kk
N
b
=
11
1
N j=b
j=
å
( )
,
0,
1
kP
l
k
=
N

deg
P
_{j}( )
l =
j
( )
nR
l
( )
( )
10,
01
R
_{}l
=
R
l
=
1,
0
n na
º
b
º
( )
nP
l
n
( )
( )
1 ,( )
{
}
0,
0,1,..., ,...,
n n n n n k k kP
l
g
R
l
c
R
l
n
M
N
=æ
ö
=
_{ç}
+
_{÷}
Î
è
å
ø
ng
c
_{n k}_{,}a b c d
_{n}, , ,
_{n} _{n} _{n} ,,
n n kg c
(
)
(
)
0 1 1 10
,
1,
2 ,
,
n n n n M n M n Ma
n M
c
M
n N
c
g
_{g}
g
g
g
g
ag
+ + +=
£ £
=
=
< £ 
=
(
)
(
)
, 1 1, 0, 1 , 1 1,0
,
0,
1 .
n n n n n n n n n nb
n M
d
M n N
c
c
c
c
c
 +  +=

£ £
=
=

< £
, (22)
where is defined by (12) and
. (23)
Since the expansion
holds, then from Eqn. (11) we have
.
Considering the Eqn. (22) we get
(24)
(25)
Eqn. (24) is important for the solution of inverse spectral problem. Firstly, are determined by using Eqn. (23) and then quantities are found from system of Eqn. (24). Thus we can find unknowns with aid of from Eqn. (25).
Lemma 7. For any fixed the system of Eqn. (24) is identically solvable. Proof. It is clear that
,
are linear independent and from the Eqn. (23), we have
( ) ( )
1 , 1 0 N n n j m j n nm n k km j j kP
R
G
G
h
_{l}
_{l}
_{g}
_{c}
b
= =é
ù
=
_{ê}
+
_{ú}
ë
û
å
å
h
( ) ( )
1 N nm n j m j j jG
h
R
l
R
l
b
==
å
( )
(j)( )
0 j j k k k Rl
w Pl
= =å
( ) ( )
11
1
,
0,
0,
N n j m j nm j j nP
l
R
l
d
n
s
n
b
hg
==
³
=
å
1 , 0 0, 0, 1, 1, n nm n k km k G c
G m n n = +å
= =  ³ 1 , 2 01
,
0, N 1.
n nn n k kn k nG
c
G
n
hg
=+
å
=
=
nmG
,,
0,
1
n kk
n
c
=
ng
c
_{n k}_{,}n
( )
1( )
2( )
1 2,
,...,
,
0,
1
n n n N n NR
R
R
v
l
l
l
n
N
b
b
b
æ
ö
=
ç
_{ç}
÷
_{÷}
=
è
ø
.
The basic determinant of the system (24)
. (26)
Lemma 8. Let be a solution of the system (24). Then
Proof. See [19].
Therefore we can determine the entries of the matrix from the formulas Eqn. (20) and Eqn. (21).
4. Inverse Problem for Two Spectra
Consider the boundary value problem
(27) with the boundary conditions
, (28)
where is defined in (3). Now, the matrix of coefficients of Eqn. (27) is denoted by which has the same form with matrix . If we delete the first row and the first column of the matrix then we have tridiagonal matrix which has number eigenvalues
. Assume that eigenvalues of matrix are distinct and real. Thus we can write
.
The solution of Eqn. (27) is denoted by provided that . It is clear that the eigenvalues are zeros of the polynomial . While we determine entries of , we will use eigenvalues of matrices and .
(
,
)
ns n sG
=
v v
{ }
1{
(
)
}
1 , 0 , 0det
_{ij} ndet
_{i},
_{j} n i j i jG
_{=}v v
==
,
0,
1
nmG
m
=
n
1 , 0 0, 0, N 1. n nm n k km k G c
G n = +å
> = J 1 1 1,
1,
1,
n n n n n n n na y
_{} _{}+
b y
+
a y
_{+}=
lr
y
n
=
N
0 N0
y
=
y
=
nr
J
_{1} J J 1 1 N ´ NJ
_{1} N1,
1,
1
kk
N
µ
=

J
1 1 2...
N 1µ µ
<
< <
µ
_{}( )
{
Q
nl
}
Q
0( )
l
=
0,
Q
1( )
l
=
1
,
1,
1
jj
N
µ
=

Q
_{N}( )
l
J JJ
_{1}Now we will give an important lemma for the inverse spectral problem according to the two spectra.
Lemma 9. The eigenvalues of matrices and alternate, i.e.
.
Proof. See [19].
Additionally, we can find the normalized numbers by aid of two spectrums and of the matrices and respectively. Assume that
, (29)
and require that . is a meromorphic function,
, (30)
its poles and zeros coincide with the eigenvalues of the problem (1)(2) and (27)(28), respectively. We see that the function satifies the equation
(31)
Now, if the equality Eqn. (31) is multiplied by and the Eqn. (6) (for ) is multiplied by then the second result is substracted from the first and sum by , we obtain:
or
,
and then for , we have
J
J
_{1} 1 1 2 2...
N 1 N 1 Nl µ l
<
<
<
µ
< <
l
<
µ
<
l
kb
1, ,...,
2 Nl l
l
µ µ
_{1}, ,...,
_{2}µ
_{N}_{1} JJ
1( )
( ) ( ) ( )
n n nf
l
=
Q
l
+
m
l
P
l
( )
0
Nf
l =
m
( )
l
( )
N_{( )}
( )
NQ
m
P
l
l
l
=
( )
nf
l
( )
( )
( )
( )
1 1 1.
n n n n n n n na f
_{} _{}l
+
b f
l
+
a f
_{+}l lr
=
f
l
( )
n kP
l
l l
=
_{k}( )
nf
l
n
(
)
( ) ( )
(
( ) ( )
( ) ( )
)
( )
( )
( )
( )
(
)
1 1 _{1} _{1} _{1} 1 1 _{1} _{1} N N _{n} _{n} _{n} _{k} _{n} _{k} _{n} k n n n k n n _{n} _{n} _{n} _{k} _{n} _{k} _{n}a
f
P
P
f
f
P
a f
P
P
f
l
l
l
l
l l
r
l
l
l
l
l
l
  _{} _{} _{} = = _{+} _{+}ì

ü
ï
ï

_{= í}
_{ý}

ï
ï
î
þ
å
å
(
)
1( ) ( )
0 1 N k n n n k n f P al l
r
l
l
= å
= kl
®
l
.
On the other hand from Lemma 1, we get
,
and from these equalities we can find
.
As a result
. (32)
Theorem 10. Let the collections to be real numbers. These collections are spectrums of the and tridiagonal almostsymmetric matrices and , respectively, it is necessary and sufficient that they are alternate as below:
.
5. Conclusion
While solving the inverse spectral problem from two spectra, firstly we determine the normalized numbers of the matrix by using the eigenvalues of the matrices and , respectively. Thus, we reduce the problem from two spectra to spectral analysis, and then we determine the values , and from the formulas (23)(25) by aid of the eigenvalues and the normalized numbers of the matrix .
( )
( )
0 N k k N ka P
Q
l
b
l
¢
=
(
) ( )
0 1 1 1( )
det
J

l
I
= 
1
Na a a c
...
_{M M}_{+}...
c P
_{N}_{} _{N}l
(
) ( )
1( )
1 1 1 1det
J

l
I
= 
1
Na a c
...
_{M M}+...
c Q
_{N} _{N}l
( ) (
1) (
)
( ) (
1) (
1)
0 1 1 1 1 1 1...
...
,
...
...
...
...
N N N N M M N M M NP
Q
a a a c
c
a a c
c
l l
l l
l µ
l µ
l
l
+  +


=
=
(
)
(
)
1 1 1 N k j j j k k N k j jl l
b
l
µ
= ¹ ==
Õ
Õ
{ }
{ }
1 1,
1 N N k k k kl
µ
= = N N´ N ´ 1 N 1 JJ
_{1} 1 1 2 2...
N 1 N 1 Nl µ l
<
<
<
µ
< <
l
_{}<
µ
_{}<
l
kb
J{ }
{ }
1 1,
1 N N k _{k} k _{k}l
µ
= = JJ
_{1} nmG
c
_{n}_{,k}g
n kl
b
_{k} JConsequently, the entries and are found from the Eqn. (20) and Eqn. (21). Thus, the matrix is reconstructed by using Parseval formula.
References
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n n n
a b c
d
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